On page 12 of An Introduction to Theoretical Fluid Dynamics, following the introduction of a material vector field $v_i(\mathbf a,t)=J_{ij}(\mathbf a,t)V_j(\mathbf a)$ the author wrote:
$$ \frac{\mathrm D \mathbf v}{\mathrm D t} = \left. \frac{\partial \mathbf v}{\partial t} \right| _ {\mathbf x} + \mathbf u \cdot \nabla \mathbf v - \mathbf v \cdot \nabla \mathbf u \equiv v_t+\mathcal L_{\mathbf u} \mathbf v = 0 $$
Question: Shouldn't the material derivative of $\mathbf v$ be the following? Where is the "extra" term with the negative sign from?
$$ \frac{\mathrm D \mathbf v}{\mathrm D t} = \left. \frac{\partial \mathbf v}{\partial t} \right| _ {\mathbf x} + \mathbf u \cdot \nabla \mathbf v $$
Update: I believe it has something to do with Eqn. (1.22) which states that
$$ \left. \frac{\partial \mathbf v}{\partial t} \right |_{\mathbf a} = \mathbf v\cdot\nabla\mathbf u $$
My confusion has been resolved. I was assuming that the material derivative of $\mathbf v$ should be equal to 0.
However, by definition,
$$\frac{\mathbf D \mathbf v}{\mathbf D t} = \left. \frac{\partial \mathbf v}{\partial t} \right|_{\mathbf a}$$
Therefore Eqn. (1.23) holds.